Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.math.BigInteger;
import java.util.Map.Entry;

import static java.lang.Math.*;

public class A extends PrintWriter {

    void run() {
        int n = nextInt(), m = nextInt();
        int[][] g = nextMatrix(n, m);

        boolean t = n > m;

        if (t) {
            int[][] r = new int[m][n];

            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    r[j][i] = g[i][j];
                }
            }

            g = r;

            n ^= m;
            m ^= n;
            n ^= m;
        }

        int[] ans = new int[n * m * 500 + 23];
        int p = 0;

        for (int i = 0; i < n; i++) {
            int cnt = 501;
            for (int j = 0; j < m; j++) {
                cnt = min(cnt, g[i][j]);
            }

            for (int j = 0; j < m; j++) {
                g[i][j] -= cnt;
            }

            while (--cnt >= 0) {
                ans[p++] = +(i + 1);
            }
        }

        for (int j = 0; j < m; j++) {
            int cnt = 501;
            for (int i = 0; i < n; i++) {
                cnt = min(cnt, g[i][j]);
            }

            for (int i = 0; i < n; i++) {
                g[i][j] -= cnt;
            }

            while (--cnt >= 0) {
                ans[p++] = -(j + 1);
            }
        }

        boolean ok = true;

        for (int i = 0; ok && i < n; i++) {
            for (int j = 0; ok && j < m; j++) {
                ok &= g[i][j] == 0;
            }
        }

        if (!ok) {
            p = -1;
        }

        println(p);
        for (int i = 0; i < p; i++) {
            int v = ans[i];

            boolean q = v < 0;

            if (q) {
                v *= -1;
            }

            if (q ^ t) {
                print("col ");
            } else {
                print("row ");
            }

            println(v);
        }

    }

    boolean skip() {
        while (hasNext()) {
            next();
        }
        return true;
    }

    int[][] nextMatrix(int n, int m) {
        int[][] matrix = new int[n][m];
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                matrix[i][j] = nextInt();
        return matrix;
    }

    String next() {
        while (!tokenizer.hasMoreTokens())
            tokenizer = new StringTokenizer(nextLine());
        return tokenizer.nextToken();
    }

    boolean hasNext() {
        while (!tokenizer.hasMoreTokens()) {
            String line = nextLine();
            if (line == null) {
                return false;
            }
            tokenizer = new StringTokenizer(line);
        }
        return true;
    }

    int[] nextArray(int n) {
        int[] array = new int[n];
        for (int i = 0; i < n; i++) {
            array[i] = nextInt();
        }
        return array;
    }

    int nextInt() {
        return Integer.parseInt(next());
    }

    long nextLong() {
        return Long.parseLong(next());
    }

    double nextDouble() {
        return Double.parseDouble(next());
    }

    String nextLine() {
        try {
            return reader.readLine();
        } catch (IOException err) {
            return null;
        }
    }

    public A(OutputStream outputStream) {
        super(outputStream);
    }

    static BufferedReader reader;
    static StringTokenizer tokenizer = new StringTokenizer("");
    static Random rnd = new Random();
    static boolean OJ;

    public static void main(String[] args) throws IOException {
        OJ = System.getProperty("ONLINE_JUDGE") != null;
        A solution = new A(System.out);
        if (OJ) {
            reader = new BufferedReader(new InputStreamReader(System.in));
            solution.run();
        } else {
            reader = new BufferedReader(new FileReader(new File(A.class.getName() + ".txt")));
            long timeout = System.currentTimeMillis();
            while (solution.hasNext()) {
                solution.run();
                solution.println();
                solution.println("----------------------------------");
            }
            solution.println("time: " + (System.currentTimeMillis() - timeout));
        }
        solution.close();
        reader.close();
    }
}